differentiate. y = ln (8x^2 - 3x + 5) y =

Explanation:

Step1: Apply chain - rule

Let $u = 8x^{2}-3x + 5$, then $y=\ln(u)$. The chain - rule states that $\frac{dy}{dx}=\frac{dy}{du}\cdot\frac{du}{dx}$. First, find $\frac{dy}{du}$. Since $y = \ln(u)$, $\frac{dy}{du}=\frac{1}{u}$.

Step2: Find $\frac{du}{dx}$

Differentiate $u = 8x^{2}-3x + 5$ with respect to $x$. Using the power rule $\frac{d}{dx}(ax^{n})=nax^{n - 1}$, we have $\frac{du}{dx}=\frac{d}{dx}(8x^{2}-3x + 5)=16x-3$.

Step3: Calculate $\frac{dy}{dx}$

By the chain - rule $\frac{dy}{dx}=\frac{dy}{du}\cdot\frac{du}{dx}$. Substitute $\frac{dy}{du}=\frac{1}{u}$ and $\frac{du}{dx}=16x - 3$ into it, and replace $u$ with $8x^{2}-3x + 5$. So $\frac{dy}{dx}=\frac{16x - 3}{8x^{2}-3x + 5}$.

Answer:

$\frac{16x - 3}{8x^{2}-3x + 5}$